* They have only one torch and the river is too risky to cross without the torch.
* If all people cross simultaneously then torch light wont be sufficient.
* Speed of each person of crossing the river is different.cross time for each person is 1 min, 2 minutes, 7 minutes and 10 minutes.

What is the shortest time needed for all four of them to cross the river ?

The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 minutes. Is that it? No. That would make this question too simple even as a warm up question.

Let's brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let's put all this together.

1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)

I bought three toys for my triplet boys (one for each). I placed the toys in the dark store. One by one each boy went to the store and pick the toy. What is the probability that no boy will choose his own toy?

There are 100 doors. 100 strangers have been gathered in the adjacent room. The first one goes and opens every door. The second one goes and shuts down all the even numbered doors – second, fourth, sixth... and so on. The third one goes and reverses the current position of every third door (third, sixth, ninth… and so on.) i.e. if the door is open, he shuts it and if the door is shut, he switches opens it. All the 100 strangers progresses in the similar fashion.

After the last person has done what he wanted, which doors will be left open and which ones will be shut at the end?

Think deeply about the door number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will open the door; pass 2, 2nd one will close it; pass 4, open; pass 7, close; pass 8, open; pass 14, close; pass 28, open; pass 56, close.
Thus we can say that the door will just end up back in its original state for each pair of divisor. But what about the cases in which the pair of divisor has analogous number for example door number 16? 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurrent because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will remain open at the end.

A pack of cards has 52 cards. You are blindfolded. Out of 52, 42 cards are facing down while 10 are facing up. You have been asked to divide this pack of cards into two decks - so that each deck contains an equal number of face up cards. Remember, you are blindfolded.

Take 10 number of cards in a new deck and change their face direction. For example- You create a new deck of 10 cards and out of 10, 3 faces up in the new deck. So remaining 7 faces up are in the old deck. But hey! while creating the new deck you reversed the face direction of new cards. So actually the 3 cards which were facing up are actually face down in the new deck while 7 faces up.

Explanation:
It's a maths magical square root series as :
Square root of number 121 is 11
Square root of number 12321 is 111
Square root of number 1234321 is 1111
Square root of number 123454321 is 11111
Square root of number 12345654321 is 111111
Square root of number 1234567654321 is 1111111
Square root of number 123456787654321 is 11111111
Square root of number 12345678987654321 is 111111111 (answer)

There can be myriad ways to create a palindrome. One day, I thought of making my own palindrome. I thought of a number and then decided to add the reversed number to it. Sadly, I did not get a palindrome.

So I kept repeating this step and eventually I succeeded in creating a palindrome. I don't know if you can always create a palindrome using this method but I was able to generate one of four digits.

Aishwarya was first to board to her flight to delhi.
She forgot her seat number and picks a random seat for herself.
After this, every single person who get to the flight sits on his seat if its available else chooses any available seat at random.
Abhishek is last to enter the flight and at that moment 99/100 seats were occupied.

one of two is the possibility
1. If any of the first 99 people sit in Abhishek seat, Abhishek will not get to sit in his own seat.
2. If any of the first 99 people sit in Aishwarya's seat, Abhishek will get to sit in his seat.

Explanation:
There can be only two cases. Either Zoe is a liar or Joe is a liar.

First Case
Let us assume that Zoe is a liar and Joe is a truthful person.
If I asked the question from Joe, the answer will be yes. If I asked the question from Zoe, the answer will be no. Thus in this case, I must have asked from Joe.

Second Case
Let us assume that Zoe is a truthful and Joe is a liar.
If I asked the question from Joe, the answer will be yes. If I asked the question from Zoe, the answer will be no. Thus in this case as well, I must have asked from Joe only.